In the figure,stackrel{leftrightarrow}{ AB }, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the tangents from point $A$ to the circle be $AB$ and $AC$. By the theorem,lengths of tangents drawn from an external point to a circle are eq

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) Let the tangents from point $A$ to the circle be $AB$ and $AC$. By the theorem,lengths of tangents drawn from an external point to a circle are equal,so $AB = AC$.Let the tangent $PQ$ touch the circle at point $M$. Then $PM = PB$ and $QM = QC$ (tangents from points $P$ and $Q$ respectively).The perimeter of $\Delta APQ = AP + PQ + AQ = AP + (PM + MQ) + AQ$.Substituting the equal tangent lengths: Perimeter $= AP + PB + QC + AQ = (AP + PB) + (AQ + QC) = AB + AC$.Since $AB = AC$,the perimeter $= 2 AB$.Given that the perimeter is $16$,we have $2 AB = 16$,which implies $AB = 8$.

Part $I$	Part $II$
$1.$ In $\Delta ABC, AB=3, BC=4, AC=5$	$a.$ In-radius $= 1$
$2.$ In $\Delta PQR, PQ=5, QR=12, PR=13$	$b.$ In-radius $= 2$
$3.$ In $\Delta XYZ, XY=8, YZ=15, XZ=17$	$c.$ In-radius $= 3$
$4.$ In $\Delta MNP, MN=20, NP=21, MP=29$	$d.$ In-radius $= 6$

In the figure,$\stackrel{\leftrightarrow}{ AB }$,$\stackrel{\leftrightarrow}{ AC }$ and $\stackrel{\leftrightarrow}{ PQ }$ are tangents to $\odot( O , r)$. If the perimeter of $\Delta APQ$ is $16$,then $AB = \ldots$

Similar Questions

If two tangents inclined at an angle $60^{\circ}$ are drawn to a circle of radius $3 \, cm$,then the length of each tangent is equal to (in $cm$):

If a rhombus $ABCD$ is inscribed in a circle with center $O$ and radius $r$,then the rhombus $ABCD$ is $\ldots \ldots$

$A$ circle touches all four sides of $\square ABCD$. If $AB = 5, BC = 8$ and $CD = 6$,then $AD = .......$

$\overline{ PA }$ and $\overline{ PB }$ are tangents to the $\odot( O , r)$ drawn from a point $P$ outside the circle,then ..... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module